Algebra Examples

$(1, 9)$ , $(1, 10)$ , $(1, 11)$

Step 1

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Step 2

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$9 = a {(1)}^{2} + b (1) + c, 10 = a {(1)}^{2} + b (1) + c, 11 = a {(1)}^{2} + b (1) + c$

Step 3

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

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Step 3.1

Solve for $a$ in $9 = a + b + c$ .

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Step 3.1.1

Rewrite the equation as $a + b + c = 9$ .

$a + b + c = 9$

$10 = a + b + c$

$11 = a + b + c$

Step 3.1.2

Move all terms not containing $a$ to the right side of the equation.

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Step 3.1.2.1

Subtract $b$ from both sides of the equation.

$a + c = 9 - b$

$10 = a + b + c$

$11 = a + b + c$

Step 3.1.2.2

Subtract $c$ from both sides of the equation.

$a = 9 - b - c$

$10 = a + b + c$

$11 = a + b + c$

$a = 9 - b - c$

$10 = a + b + c$

$11 = a + b + c$

$a = 9 - b - c$

$10 = a + b + c$

$11 = a + b + c$

Step 3.2

Replace all occurrences of $a$ with $9 - b - c$ in each equation.

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Step 3.2.1

Replace all occurrences of $a$ in $10 = a + b + c$ with $9 - b - c$ .

$10 = (9 - b - c) + b + c$

$a = 9 - b - c$

$11 = a + b + c$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Combine the opposite terms in $(9 - b - c) + b + c$ .

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Step 3.2.2.1.1

Add $- b$ and $b$ .

$10 = 9 - c + 0 + c$

$a = 9 - b - c$

$11 = a + b + c$

Step 3.2.2.1.2

Add $9 - c$ and $0$ .

$10 = 9 - c + c$

$a = 9 - b - c$

$11 = a + b + c$

Step 3.2.2.1.3

Add $- c$ and $c$ .

$10 = 9 + 0$

$a = 9 - b - c$

$11 = a + b + c$

Step 3.2.2.1.4

Add $9$ and $0$ .

$10 = 9$

$a = 9 - b - c$

$11 = a + b + c$

$10 = 9$

$a = 9 - b - c$

$11 = a + b + c$

$10 = 9$

$a = 9 - b - c$

$11 = a + b + c$

Step 3.2.3

Replace all occurrences of $a$ in $11 = a + b + c$ with $9 - b - c$ .

$11 = (9 - b - c) + b + c$

$10 = 9$

$a = 9 - b - c$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Combine the opposite terms in $(9 - b - c) + b + c$ .

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Step 3.2.4.1.1

Add $- b$ and $b$ .

$11 = 9 - c + 0 + c$

$10 = 9$

$a = 9 - b - c$

Step 3.2.4.1.2

Add $9 - c$ and $0$ .

$11 = 9 - c + c$

$10 = 9$

$a = 9 - b - c$

Step 3.2.4.1.3

Add $- c$ and $c$ .

$11 = 9 + 0$

$10 = 9$

$a = 9 - b - c$

Step 3.2.4.1.4

Add $9$ and $0$ .

$11 = 9$

$10 = 9$

$a = 9 - b - c$

$11 = 9$

$10 = 9$

$a = 9 - b - c$

$11 = 9$

$10 = 9$

$a = 9 - b - c$

$11 = 9$

$10 = 9$

$a = 9 - b - c$

Step 3.3

Since $11 = 9$ is not true, there is no solution.

No solution